Induction over prime numbers

This was a question which intrigued me when I was a student: is there a meaningful mathematical statement about finite fields which is proven by induction on the characteristic of a field?

Finally and many years later I got a partial answer: a meaningful statement about prime numbers proven by induction on the size of prime number in question.

This is a note A Lemma on Divisibility by Peter Walker in a recent issue of The American Mathematical Monthly 115 no. 4 (2008 ) p. 338:

(*) For all primes , implies or .

His proof uses division with remainder but not Euclidean algorithm for finding the greatest common divisor of two integers.

We say that a prime is genuine if it satisfies (*) for all integers and . We shall prove by induction on that all primes are genuine.

Basis of induction. is a genuine prime number. Indeed if a product of two integers is even then at least one of the multiplicands is even.

Inductive step. Let be the least non-genuine prime number so that but does not divide and does not divide .

Using division with remainder, we can write and where . Of course, . If either or then divides or as required. If not, then both and are at least and can be factorised into primes: and (existence of factorisation can be proved earlier). Now and for some integer we have wher all , are less than and are therefore genuine prime numbers. Since none of , can divide , they divide and can be cancelled out one by one from the equation, leaving an obviously contradictory equality .

How do you know that ab might not be both odd and even, without deriving that fact from unique factorization? Well, one answer is that if m and n are integers for which 2m+1 = 2n, then 2(m-n) = 1, a case of the “contradictory inequality vp = 1” that you mentioned. By the way, the impossibility of vp = 1 unless v = p = 1 (for positive integers) follows easily from the Peano axioms and the recursive definition of multiplication.

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@Charles: You know it from induction “step”. Contrary to popular opinion, induction with total history (if all primes less than p are genuine, then p is genuine) _doesn’t_ need a basis, since it is just a special case of a step: all primes less than 2 _are_ surely genuine (since there are none), so 2 is genuine by induction step.

Only inductions with bounded history (eg, if P(k-1) and P(k), then P(k+1)) need a basis, since you can’t conclude P(0) by the step if P(-1) doesn’t hold or even make sense in the context.

@Alexandre: So the real proof is even shorter, in fact, and the whole even/odd distinction is a special case of what’s going on in the step. Really, in the step we invoke theorem about division with remainder, which obviously answers second Charles’s question as well (when p is 2, remainder can be 0 or 1, and can’t be both).